Question

Write a Pythagorean triplet whose smallest number is
(i) 6 (ii) 14 (iii) 16 (iv) 20

Answer 1

We know that $2m,m^2-1$ and $m^2+1$ is a Pythagorean triplet where m>1
(i) One number=6
$\therefore 2m=6\Rightarrow m=3$ $\therefore$ Other members of triplet will be
$m^2-1=(3{)}^2-1=9-1=8$
and $m^2+1=(3{)}^2+1=9+1=10$
$\therefore$ Pythagorean triplet is 6, 8, 10
(ii) Let $2m=14\Rightarrow m=\frac{14}{2}=7$
$\therefore m^2-1=(7{)}^2-1=49-1=48$
and $m^2+1=(7{)}^2+1=49+1=50$
(iii) Let $2m=16\Rightarrow m=8$
$\therefore m^2-1=(8{)}^2-1=64-1=63$
and $m^2+1=(8{)}^2+1=64+1=65$
$\therefore$ Pythagorean triplet= 16, 63, 65
(iv) Let $2m=20\Rightarrow m=10$
$\therefore m^2-1=(10{)}^2-1=100-1=99$
and $m^2+1=(10{)}^2+1=100+1=101$
$\therefore$ Pythagorean triplet = 20, 99, 101